JEE Main 2017 · 8 Apr · Shift 1 · Physics Q5

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PhysicsSingle correct+4 / -1Question 5 of 90

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :

A disc of mass M and radius R mounted on an axle fixed to a wall; a string wound on its rim hangs down to a block of mass m.

\dfrac{2mg}{2m+M}

\dfrac{2Mg}{2m+M}

\dfrac{2mg}{2M+m}

\dfrac{2Mg}{2M+m}

Answer and worked solution

Official answer: A

For the block, $mg-T=ma$ . For the disc, $TR=\frac12MR^2\cdot\frac aR$ , so $T=\frac12Ma$ . Then
$a=\frac{mg}{m+M/2}=\frac{2mg}{2m+M}.$

Topics:Torque and rotational dynamics Moment of inertia

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 5